Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $p = \dfrac{-3r + 9}{-5r^2 + 55r - 120} \times \dfrac{r^2 - 7r - 8}{-2r - 10} $
Answer: First factor out any common factors. $p = \dfrac{-3(r - 3)}{-5(r^2 - 11r + 24)} \times \dfrac{r^2 - 7r - 8}{-2(r + 5)} $ Then factor the quadratic expressions. $p = \dfrac {-3(r - 3)} {-5(r - 8)(r - 3)} \times \dfrac {(r - 8)(r + 1)} {-2(r + 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-3(r - 3) \times (r - 8)(r + 1) } { -5(r - 8)(r - 3) \times -2(r + 5)} $ $p = \dfrac {-3(r - 8)(r + 1)(r - 3)} {10(r - 8)(r - 3)(r + 5)} $ Notice that $(r - 8)$ and $(r - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-3\cancel{(r - 8)}(r + 1)(r - 3)} {10\cancel{(r - 8)}(r - 3)(r + 5)} $ We are dividing by $r - 8$ , so $r - 8 \neq 0$ Therefore, $r \neq 8$ $p = \dfrac {-3\cancel{(r - 8)}(r + 1)\cancel{(r - 3)}} {10\cancel{(r - 8)}\cancel{(r - 3)}(r + 5)} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $p = \dfrac {-3(r + 1)} {10(r + 5)} $ $ p = \dfrac{-3(r + 1)}{10(r + 5)}; r \neq 8; r \neq 3 $